I don’t understand the variance of the binomial

Question 1

I feel really dumb even asking such a basic question but here goes:

If I have a random variable $X$ that can take values $0$ and $1$ , with $P(X=1) = p$ and $P(X=0) = 1-p$ , then if I draw $n$ samples out of it, I’ll get a binomial distribution.

The mean of the distribution is

$\mu = np = E(X)$

The variance of the distribution is

$\sigma^2 = np(1-p)$

Here is where my trouble begins:

Variance is defined by $\sigma^2 = E(X^2) - E(X)^2$ . Because the square of the two possible $X$ outcomes don’t change anything ( $0^2 = 0$ and $1^2 = 1$ ), that means $E(X^2) = E(X)$ , so that means

$\sigma^2 = E(X^2) - E(X)^2 = E(X) - E(X)^2 = np - n^2p^2 = np(1-np) \neq np(1-p)$

Where does the extra $n$ go? As you can probably tell I am not very good at stats so please don’t use complicated terminology :s

Question 2

A random variable $X$ taking values $0$ and $1$ with probabilities $P(X=1)=p$ and $P(X=0)=1-p$ is called a Bernoulli random variable with parameter $p$ . This random variable has
$\begin{eqnarray*} E(X)&=&0\cdot (1-p) + 1\cdot p = p\\ E(X^2)&=&0^2\cdot(1-p) + 1^2\cdot p = p\\ Var(X)&=& E(X^2)-(E(X))^2=p-p^2=p(1-p) \end{eqnarray*}$
Suppose you have a random sample $X_{1},X_{2},\cdots,X_{n}$ of size $n$ from $Bernoulli(p)$ , and define a new random variable $Y=X_{1}+X_{2}+\cdots +X_{n}$ , then the distribution of $Y$ is called Binomial, whose parameters are $n$ and $p$ .
The mean and variance of the Binomial random variable Y is given by
$\begin{eqnarray*} E(Y)&=&E(X_{1}+X_{2}+\cdots + X_{n})=\underbrace{ p+p+\cdots +p}_{n}=np\\ Var(Y)&=& Var(X_{1}+X_{2}+\cdots + X_{n})=Var(X_{1})+Var(X_{2})+\cdots + Var(X_{n})\\ & &\text{ (as $X_{i}$'s are independent)} \\ &=&\underbrace{p(1-p)+p(1-p)+\cdots+ p(1-p)}_{n}\quad \text{ (as $X_{i}$'s are identically distributed)} \\ &=&np(1-p) \end{eqnarray*}$

I don’t understand the variance of the binomial

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